EMF of A Galvanic Cell

Every galvanic or voltaic cell is made up of two half-cells, the oxidation half-cell (anode) and the reduction half-cell (cathode). The potentials of these half-cells are always dif­ferent. On account of this difference in electrode potentials, the electric current moves from the electrode at higher potential to the electrode at lower potential, i.e., from cathode to anode. The direction of the flow of electrons is from anode to cathode.

The difference in potentials of the two half-cells is known as the electromotive force (EMF) of the cell or cell potential.

The emf of the cell or cell potential can be calculated from the values of electrode potentials of the two half-cells constitut­ing the cell. The following three methods are in use:

When oxidation potential of anode and reduction poten­tial of cathode are taken into account:

Difference between EMF and Potential Difference

The potential difference is the difference between the electrode potentials of the two electrodes of the cell under any condition while emf is the potential generated by a cell when there is zero electron flow, i.e., it draws no current. The points of difference are given below:

S.No.

EMF

Potential Difference

1

t is the potential difference between two electrodes when no current is flowing in the circuit.

It is the difference of the electrode potentials of the two electrons two electrodes when the cell is under operation.

2

It is the maximum voltage that the cell can deliver.

It is always less than the maximum the cell can deliver.

3

It is responsible for the steady flow of current in the cell.

It is not responsible for the steady flow of current in the cell.

Nernst Equation

The electrode potential and the emf of the cell depend upon the nature of the electrode, temperature and the activities (concentrations) of the ions in solution. The variation of electrode and cell potentials with concentration of ions in solution can be obtained from thermodynamic considerations. For a general reaction such as

M1A + m2B ..... n1X + n2Y + .... .......(i)

Occurring in the cell, the Gibbs free energy change is given by the equation

where

'a' represents the activities of reactants and products under a given set of conditions and

?Go refers to free energy change for the reaction when the various reactants and products are present at standard conditions.

The free energy change of a cell reaction is related to the electrical work that can be obtained from the cell, i.e.,

?Go = -nFEcell and ?Go = -nFEo.

On substituting these values in Eq. (ii) we get

This equation is known as Nernst equation.

Putting the values of

R=8.314 JK-1 mol-1,

T = 298 K and

F=96500 C,

Eq. (iv) reduces to

Potential of Single Electrode (Anode)

Consider the general oxidation reaction,

M → Mn+ + ne-

Applying Nernst equation,

where Eox is the oxidation potential of the electrode (anode), is the standard oxidation potential of the electrode.

Let us consider a Daniell cell to explain the above equations. The concentrations of the electrolytes are not 1 M.

Zn(s)+Cu2+(aq) Zn2+(aq) + Cu(s)

Cell Representation: Zn(s)|Zn2+(aq)||Cu2+(aq)|Cu

Potential at zinc electrode (Anode)

Potential at Copper Electrode (Cathode)

Emf of the cell

Ecell = Eox + Ered

The value of n = 2 for both zinc and copper.

Let us consider an example, in which the values of n for the two ions in the two half-cells are not same. For example, in the cell

Cu|Cu2+||Ag+|Ag

The cell reaction is

Cu(s) + 2Ag+ → Cu2+ + 2Ag

The two half-cell reaction are:

Cu → Cu2+ + 2e-

Ag+ + e- → Ag

The second equation is multiplied by 2 to balance the number of electrons.

2Ag+ + 2e- → 2 Ag

Ecell = Eox + Ered

All the EMFs that have been used are reduction potentials in volts. Whatever be the value of reduction potentials, substitute them as such, taking into account the signs.